当我尝试在 glmm 中包含线性变量时，Summary() 返回 NaN 值

Question

I am attempting to run a model using glmmTMB . When I include avgt60, it does weird things in the model and I am not really sure why. When I include it as a non poly term, it gives me NaN values. When I include it as a poly() term, it throws the entire model off. When I exclude it, it seems to be fine... I am new to this type of work so any advice is appreciated!

m1 <- glmmTMB(dsi ~ poly(rh60, degree = 2) + poly(wndspd60, degree = 2) + poly(raintt60, degree = 2) + avgt60 + (1|year) + (1|site),
              family = "nbinom2", data = weather1)

I get:

Family: nbinom2  ( log )
Formula:          dsi ~ poly(rh60, degree = 2) + poly(wndspd60, degree = 2) + poly(raintt60,      degree = 2) + avgt60 + (1 | year) + (1 | site)
Data: weather1

     AIC      BIC   logLik deviance df.resid 
  1647.9   1687.9   -813.0   1625.9      269 

Random effects:

Conditional model:
 Groups Name        Variance  Std.Dev. 
 year   (Intercept) 5.883e-24 2.426e-12
 site   (Intercept) 6.396e-07 7.997e-04
Number of obs: 280, groups:  year, 3; site, 6

Dispersion parameter for nbinom2 family (): 0.232 

Conditional model:
                            Estimate Std. Error z value Pr(>|z|)
(Intercept)                  -7.8560        NaN     NaN      NaN
poly(rh60, degree = 2)1      47.9631        NaN     NaN      NaN
poly(rh60, degree = 2)2      -5.4370        NaN     NaN      NaN
poly(wndspd60, degree = 2)1  61.7092        NaN     NaN      NaN
poly(wndspd60, degree = 2)2 -74.9432        NaN     NaN      NaN
poly(raintt60, degree = 2)1  27.2669        NaN     NaN      NaN
poly(raintt60, degree = 2)2 -72.9072        NaN     NaN      NaN
avgt60                        0.4384        NaN     NaN      NaN

But, without the avgt60 variable...

m1 <- glmmTMB(dsi ~ poly(rh60, degree = 2) + poly(wndspd60, degree = 2) + poly(raintt60, degree = 2) + (1|year) + (1|site),
              family = "nbinom2", data = weather1)


 Family: nbinom2  ( log )
Formula:          dsi ~ poly(rh60, degree = 2) + poly(wndspd60, degree = 2) + poly(raintt60,      degree = 2) + (1 | year) + (1 | site)
Data: weather1

     AIC      BIC   logLik deviance df.resid 
  1648.2   1684.6   -814.1   1628.2      270 

Random effects:

Conditional model:
 Groups Name        Variance  Std.Dev. 
 year   (Intercept) 2.052e-10 1.433e-05
 site   (Intercept) 4.007e-10 2.002e-05
Number of obs: 280, groups:  year, 3; site, 6

Dispersion parameter for nbinom2 family (): 0.23 

Conditional model:
                            Estimate Std. Error z value Pr(>|z|)    
(Intercept)                   1.3677     0.3482   3.928 8.56e-05 ***
poly(rh60, degree = 2)1      23.8058     9.6832   2.458 0.013953 *  
poly(rh60, degree = 2)2      -0.3452     4.2197  -0.082 0.934810    
poly(wndspd60, degree = 2)1  34.4332    10.1328   3.398 0.000678 ***
poly(wndspd60, degree = 2)2 -61.2044     6.5179  -9.390  < 2e-16 ***
poly(raintt60, degree = 2)1  12.0109     6.4949   1.849 0.064417 .  
poly(raintt60, degree = 2)2 -57.2197     6.0502  -9.457  < 2e-16 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

If I leave avgt60 in as a poly() term, it throws the entire model off, and nothing is significant. Any thoughts here?

Here's a link to the dataset, with site names redacted: https://docs.google.com/spreadsheets/d/1mFDK_YEshvgGPHpvqu4o6TbFfwKRgHaVUVGOIZnsq7c/edit?usp=sharing

Answer 1

You have 280 rows in your data set, but only 10 unique values of the predictor variables:

nrow(unique(subset(weather1, select = -c(dsi))))

This determines how complex a model you can actually fit.

You are trying to estimate 8 fixed-effect parameters ( length(fixef(m1)$cond) or ncol(model.matrix(m1)) ), two random-effect parameters (among-site and among-year variances), and one dispersion parameter (for the negative binomial parameter) = 11 (or length(m1$fit$par) ). This is more parameters than you have unique predictor combinations!

Murtaugh (2007) makes the point that when you have a nested design (values of the predictor variables change only between groups, not within groups) you will get the same effects estimated if you aggregate the response variable for each group (or site/year combination in your case) to its mean. (If you have unbalanced groups as in this case you need to fit a model with weights, and this approach doesn't work well for non-Gaussian responses, but the principle is similar.)

If you leave out avgt60 you "only" have 10 parameters. I still don't trust this model very much, it's badly overparameterized (normally you're aiming for something like (# observations)/(# data points) at least 10, preferably 20 ...) To be honest I'm not even sure why it's working - I think because the site and year variances are basically collapsing to zero and removing themselves from the model, so you "only" have 8 parameters to estimate?

Here's what the data look like:

• dsi values for sites 5 and 6 are always zero (only measured in 2021)
• dsi values are very high in 2019, only two sites (1 and 3) measured
• there is no particular pattern, and certainly not one that's not confounded with site and year.

I would probably try to draw only qualitative conclusions, or very simple quantitative conclusions, from these data ...

library(tidyverse); theme_set(theme_bw())
w3 <- (weather1
    |> as_tibble()
    |> select(-date)
    |> pivot_longer(-c(site, year, dsi), names_to = "var")
    |> mutate(across(c(year,site), factor))
)

theme_set(theme_bw(base_size = 20)  + theme(panel.spacing = grid::unit(0, "lines")))
(ggplot(w3)
    + aes(x = value, y = dsi, colour = site, shape = year)
    + stat_sum(alpha = 0.6)
    + stat_summary(fun = mean)
    + stat_summary(fun = mean, geom = "line", aes(group = 1), colour = "gray")
    + facet_wrap(~var, scale = "free_x")
    + scale_y_sqrt()
)

Murtaugh, Paul A. “Simplicity and Complexity in Ecological Data Analysis.” Ecology 88, no. 1 (2007): 56–62.